# Solution to the Four Squares Inside a Rectangle Puzzle +-- {.image} [[FourSquaresInsideaRectangle.png:pic]] > Four squares inside a rectangle. What’s the angle? =-- ## Solution by [[Angles in a triangle]], [[Angles at a point on a straight line]], and [[Angles in parallel lines]] +-- {.image} [[FourSquaresInsideaRectangleLabelled.png:pic]] =-- In the above diagram, $A$ is horizontally across from $F$ and $D$ is vertically below. So the quadrilateral $A C D F$ has all four interior angles $90^\circ$ and so is a rectangle. Considering angles at $B$, and using that $F B$ is a diagonal of the left-hand square, we have: $$ C \hat{B} E + 45^\circ + F \hat{B} A = 180^\circ $$ Over at $E$, we have: $$ F \hat{E} D = 45^\circ + B \hat{E} C $$ Then as triangle $B C E$ is a [[right-angled triangle]]: $$ B \hat{E} C + C \hat{B} E = 90^\circ $$ Putting those together, $$ F \hat{E} D = 45^\circ + B \hat{E} C = 45^\circ + 90^\circ - C \hat{B} E = F \hat{B} A $$ So triangles $F A B$ and $F D E$ are [[similar]], and the lengths of $F B$ and $F E$ are the same, so they are [[congruent]]. This means that the lengths $F A$ and $F D$ are also the same and so the quadrilateral $A C D F$ is actually a square. The line $C F$ is therefore at $45^\circ$ to $C D$ and so the requested angle is also $45^\circ$ by [[angles in parallel lines]]. ## Solution by [[Invariance Principle]] The squares can be drawn at any angle inside the rectangle, so they can be drawn to align with the sides of the outer rectangle. This brings $B$ to $C$ and the left-hand square sits in the lower left corner of the rectangle. The line $C F$ is then evidently at $45^\circ$ to the base of the rectangle.